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2 years ago

Draw structures from the following names, and determine which compounds are optically active:(b) 3-chloro-3-methylpentane

Chemistry

1 answer:

VladimirAG [237]2 years ago

8 0

The given compound 3-chloro-3-methylpentane is not optically active .

Because this compound have plane of symmetry (POS) and center of symmetry (COS) i.e. have di-symmetry . And also forms superimposable mirror image . the compound is not optically active .

It does not have chiral center.

Here the carbon atom has only 3 distinct groups i.e. ethyl , chlorine , methyl .

<h3>What is di-symmetry?</h3>

Di-symmetry is that which have no center of symmetry and plane of symmetry and alternate axis of symmetry .

<h3>Chiral center :</h3>

Have Sp3 hybridized center (4sigma bond ) .

4 distinct group is attached to the chiral atom. form non -superimposable mirror image .

<h3>What is optical isomerism ?</h3>

Same molecular formula and same structural formula . also have same physical and chemical properties .

They differ in their behavior towards plane polarized light (ppl) .

Learn more about chiral center here:

brainly.com/question/9522537

#SPJ4

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<h3>Percent compositions:</h3>

In chemistry, percent compositions are used for us to know the relative amount of a specific element in a compound. In order to do so for hydrogen, we use the following formula, which can also be applied to any other element in a given compound:

$\% H=\frac{m_H*1}{M_{HF}}*100\%$

Where $m_H$ stands for the atomic mass of hydrogen and $M_{HF}$ for the molar mass of hydrofluoric acid. In such a way, we plug in the atomic masses of hydrogen (1.01 g/mol) and fluorine (19.0 g/mol) to obtain:

$\% H=\frac{1.01g/mol*1}{1.01g/mol+19.0g/mol}*100\%\\\\\% H=5\%$

Learn more about percent compositions: brainly.com/question/12247957

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